Worked Solutions
Introduction to Calculus — Worked Solutions (Preliminary Maths Advanced)
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Worked examples for Preliminary Maths Advanced introduction to calculus. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Differentiation from first principles
Question
Use first principles to find the derivative of $f(x) = x^2 + 3x$.
Solution
Apply the definition $f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$.
$f(x+h) = (x+h)^2 + 3(x+h) = x^2 + 2xh + h^2 + 3x + 3h$.
Subtract $f(x)$: $f(x+h) - f(x) = 2xh + h^2 + 3h$.
Divide by $h$: $2x + h + 3$. Let $h \to 0$: $f'(x) = 2x + 3$.
Write the limit at every line until you cancel the $h$ — examiners want to see it, not just the final answer.
First principles means going back to the definition of the derivative as a limit: $f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$. The idea is the gradient of a chord as the two points slide together.
Start by expanding $f(x+h)$: $(x+h)^2 + 3(x+h) = x^2 + 2xh + h^2 + 3x + 3h$.
Subtracting the original $f(x) = x^2 + 3x$ cancels the $x^2$ and $3x$, leaving $2xh + h^2 + 3h$. Notice every term has a factor of $h$ — that's what lets us divide safely.
Dividing by $h$ gives $2x + h + 3$. Finally we let $h \to 0$, and the lone $h$ disappears, so $f'(x) = 2x + 3$. That cancellation of $h$ is the whole point: it removes the $0/0$ problem.
Definition: $f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$.
- $f(x+h) = x^2 + 2xh + h^2 + 3x + 3h$
- $f(x+h) - f(x) = 2xh + h^2 + 3h$
- $\div h$: $2x + h + 3$
- $h \to 0$: $f'(x) = 2x + 3$
Where the marks go
- 1 mark: States the first-principles limit and expands $f(x+h)$
- 1 mark: Simplifies the difference quotient to $2x + h + 3$
- 1 mark: Takes the limit to get $f'(x) = 2x + 3$
Key idea
First principles uses $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$; expand, cancel the common factor of $h$, then let $h \to 0$.
Example 2 — Differentiation rules
Question
For $y = 2x^3 - 5x^2 + 4$, find $\dfrac{dy}{dx}$ and hence the gradient of the tangent at $x = 2$.
Solution
Differentiate term by term with the power rule, $\frac{d}{dx}(ax^n) = nax^{n-1}$.
$\frac{dy}{dx} = 6x^2 - 10x + 0 = 6x^2 - 10x$.
Now substitute $x = 2$: $6(4) - 10(2) = 24 - 20 = 4$.
Gradient $= 4$. The constant $4$ differentiates to $0$ — don't carry it through.
The gradient function is the derivative, so differentiate first using the power rule: multiply by the index, then drop the index by one.
Term by term: $2x^3 \to 6x^2$, $-5x^2 \to -10x$, and the constant $4 \to 0$ (a flat line has no slope). So $\frac{dy}{dx} = 6x^2 - 10x$.
The gradient at a point means we evaluate this derivative there. At $x = 2$: $6(2)^2 - 10(2) = 6 \times 4 - 20 = 24 - 20 = 4$.
So the tangent's gradient at $x = 2$ is $4$. The two-step pattern — differentiate to get the gradient function, then substitute the $x$-value — comes up constantly.
Power rule term by term.
- $\frac{dy}{dx} = 6x^2 - 10x$
Evaluate at $x = 2$.
- $6(2)^2 - 10(2)$
- $= 24 - 20$
- $= 4$
Gradient $= 4$.
Where the marks go
- 1 mark: Differentiates the cubic and quadratic terms correctly
- 1 mark: Constant differentiates to $0$, giving $\frac{dy}{dx} = 6x^2 - 10x$
- 1 mark: Substitutes $x = 2$ into the derivative
- 1 mark: Correct gradient $4$
Key idea
The derivative is the gradient function; evaluate it at a particular $x$ to get the gradient of the tangent there.